An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Explanation

Here concave mirror
`therefore` Focal length `f = - (R)/(2) = -(15)/(2) =-75. cm`
(i) Object distance u = -10 cm
`therefore` Mirror euqation `1/f = 1/u + 1/v`
`therefore 1/v = 1/f -1/u = 1/(-7.5) - (1)/(-10)`
`= - (1)/(7.5) + (1)/(10) = - (2.5)/(75)`
`therefore (1)/(v) = - (1)/(30)`
`therefore v = - 30 cm`
`therefore` Image is obtained at 30 cm from mirror,
`rArr` Magnification `m= - (v)/(u) = - (-30)/(-10) = -3`
As it is negative, image will be real, inverted and enlarged.
(ii) Object distance u = -5 cm
`therefore` Mirror euqation,
`(1)/(f) = (1)/(u) + (1)/(v) rArr 1/v -1/f -1/u`
`therefore 1/v = (1)/(-7.5) - (1)/(-5) = (1)/(7.5) + (1)/(5)`
`therefore 1/v = (-1 + 1.5)/(7.5) = (1)/(5)`
`therefore` v = 15 cm
Thus, image is positive hence it is obtained behind the mirror at 15 cm from it.
`rArr` Magnification ` m = - (v)/(u) = - (15)/(-5) = 3`
Thus , image is virtual erect and enlarged.